# How do you graph 2(y-1) > 3(x+1)?

May 13, 2015

The inequality can be rewritten under the form

$y > \frac{3}{2} x + \frac{5}{2}$

after the following operations:

$2 y - 2 > 3 x + 3$ <=> $2 y > 3 x + 5$ <=> $y > \frac{3}{2} x + \frac{5}{2}$

The graph of the equation $y = \frac{3}{2} x + \frac{5}{2}$ is a straight line. The graph of the inequality $y > \frac{3}{2} x + \frac{5}{2}$ will be formed by all the points of the plane whose coordinates $y$ have values higher than $\frac{3}{2} x + \frac{5}{2}$

Therefore, the graph of our inequality will be the semi-plane situated above the blue line defined by the equation $y = \frac{3}{2} x + \frac{5}{2}$

graph{y > 3/2 x + 5/2 [-10, 10, -5, 5]}

May 13, 2015

Distribute the $2$ and the $3$ .

$2 \left(y - 1\right) > 3 \left(x + 1\right)$

$2 y - 2 > 3 x + 3$

Convert the inequality to slope-intercept form while keeping the inequality.

Add $- 2$ to both sides.

$2 y > 3 x + 3 + 2$ =

$2 y > 3 x + 5$

Divide both sides by $2$.

$\frac{\cancel{2} y}{\cancel{2}} > \frac{3 x}{2} + \frac{5}{2}$ =

$y > \frac{3 x}{2} + \frac{5}{2}$

In order to graph $y > \frac{3 x}{2} + \frac{5}{2}$, do the following:

Determine two points on the line by temporarily making $y = \frac{3 x}{2} + \frac{5}{2}$ .

If $x = 1$; $y = \frac{\left(3\right) \cdot \left(1\right)}{2} + \frac{5}{2} = \frac{3}{2} + \frac{5}{2} = {\cancel{8}}^{4} / {\cancel{2}}^{1} = 4$
Point = $\left(1 , 4\right)$

If $x = 3$; $y = \frac{\left(3\right) \cdot \left(3\right)}{2} + \frac{5}{2} = \frac{9}{2} + \frac{5}{2} = {\cancel{14}}^{7} / {\cancel{2}}^{1} = 7$
Point = $\left(3 , 7\right)$

Plot the two points and draw a dashed line -------- through the two points. This represents that the points on the line are not part of the inequality. Then shade in the area above the dashed line.

graph{y>(3x)/2+(5)/2 [-16.02, 16, -8.01, 8.01]}